If n - 10- - x - 50- - y - - 30- - x2 - 290- - y2 - 300- - xy - -115- then find the coefficient of correlation-

Given that: n = 10 ∑ x = 50, ∑ y = 30 ∑ x^2 = 290, ∑ y^2 = 300 , ∑ xy = 115 Formula : = n∑ xy ∑ x . ∑ y√(n ∑ x^2 (∑ x)^2).√(n.∑ y^ ...

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Standard XII

Mathematics

Coefficient of Variation

If n = 10, ...

Question

If $n = 10, \sum x = 50, \sum y = - 30, \sum x^{2} = 290, \sum y^{2} = 300, \sum x y = - 115$ , then find the coefficient of correlation.

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Solution

Given that: $n = 10$
$\sum x = 50, \sum y = - 30$
$\sum x^{2} = 290, \sum y^{2} = 300$ ,
$\sum x y = - 115$
Formula :
$= \frac{n \sum x y - \sum x . \sum y}{\sqrt{n \sum x^{2} - (\sum x)^{2}} . \sqrt{n . \sum y^{2} - (\sum y)^{2}}}$
$r = \frac{[10 \times (- 115)] - [50 \times (- 30)]}{\sqrt{10 \times 290 - (50)^{2}} . \sqrt{10 \times 300 - (- 30)^{2}}}$
$= \frac{- 1150 + 1500}{\sqrt{2900 - 2500} . \sqrt{3000 - 900}}$
$= \frac{350}{\sqrt{400} . \sqrt{2100}}$
$= \frac{350}{20 \times 10 \sqrt{21}} = \frac{7}{4 \sqrt{21}}$
$= 0.381$ .

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If n=10,∑x=50,∑y=−30,∑x2=290,∑y2=300,∑xy=−115, then find the coefficient of correlation.

Given that: n=10∑x=50,∑y=−30∑x2=290,∑y2=300,∑xy=−115Formula :=n∑xy−∑x.∑y√n∑x2−(∑x)2.√n.∑y2−(∑y)2r=[10×(−115)]−[50×(−30)]√10×290−(50)2.√10×300−(−30)2=−1150+1500√2900−2500.√3000−900=350√400.√2100=35020×10√21=74√21=0.381.

If $n = 10, \sum x = 50, \sum y = - 30, \sum x^{2} = 290, \sum y^{2} = 300, \sum x y = - 115$ , then find the coefficient of correlation.